Answer:
A
Explanation:
Hydrocarbons with short chain lengths are more volatile than those with longer chains. A practical example of this can be seen in the first few members of the alkane series. They are mostly gaseous in nature and this is quite a contrast to the next few members which are solid in nature.
As we move down the group, we can see that there is an increase in the number of solids. Hence, as we go down the group we can see a relative increase in order and thus we expect more stability at room temperature compared to the volatility of the shorter chain
Answer:
its false ...............
Answer:
Percent by mass of water is 56%
Explanation:
First of all calculate the mass of hydrated compound as,
Mass of Sodium = Na × 2 = 22.99 × 1 = 45.98 g
Mass of Sulfur = S × 1 = 32.06 × 1 = 32.06 g
Mass of Oxygen = O × 14 = 16 × 14 = 224 g
Mass of Hydrogen = H × 20 = 1.01 × 20 = 20.2 g
Mass of Na₂S0₄.10H₂O = 322.24 g
Secondly, calculate mass of water present in hydrated compound. For this one should look for the coefficient present before H₂O in molecular formula of hydrated compound. In this case the coefficient is 10, so the mass of water is...
Mass of water = 10 × 18.02
Mass of water = 180.2 g
Now, we will apply following formula to find percent of water in hydrated compound,
%H₂O = Mass of H₂O / Mass of Hydrated Compound × 100
Putting values,
%H₂O = 180.2 g / 322.24 g × 100
%H₂O = 55.92 % ≈ 56%
Answer:
because they change
Explanation:
It was based on theories and discoveries
<u>Answer:</u> The average atomic mass of X is 28.09 amu
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
Mass of isotope 1 = 27.979 amu
Percentage abundance of isotope 1 = 92.21 %
Fractional abundance of isotope 1 = 0.9212
Mass of isotope 2 = 28.976 amu
Percentage abundance of isotope 2 = 4.70 %
Fractional abundance of isotope 2 = 0.0470
Mass of isotope 3 = 29.974 amu
Percentage abundance of isotope 3 = 3.09 %
Fractional abundance of isotope 3 = 0.0309
Putting values in equation 1, we get:
![\text{Average atomic mass of X}=[(27.979\times 0.9212)+(28.976\times 0.0470)+(29.974\times 0.0309)]](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20atomic%20mass%20of%20X%7D%3D%5B%2827.979%5Ctimes%200.9212%29%2B%2828.976%5Ctimes%200.0470%29%2B%2829.974%5Ctimes%200.0309%29%5D)
Hence, the average atomic mass of X is 28.09 amu
